Chemistry Homework Answer Key #5
Chapter 10 - Acid-Base Equilibria
13. pH = -log [H3O+] = -log (2.0 ´ 10-4) = 3.70 (I know, this was an insulting question, can you forgive me?)
17. Kw = [H3O+] [OH-]
[H3O+] = 10-pH = 10-8.00 = 1.00 ´ 10-8
21.
a) The fact that ephedrine acts as a base in water means that it will accept a proton from water as shown in the equation:C10H15ON (aq) + H2O (l) « C10H15ONH+ (aq) + OH- (aq)
b) For any conjugate acid/base pair, the following is true: KaKb = Kw. Therefore,
c) The Kb for ammonia is 1.79 ´ 10-5 (from the Ka for its conjugate acid, ammonium, listed on page 323); this is smaller than the Kb for ephedrine; thus, ephedrine is the stronger weak base.
23. The reaction,
HClO2 (aq) + NO2- (aq) « HNO2 (aq) + ClO2- (aq)
is the sum of the following two reactions:
HClO2 (aq) + H2O (l) « ClO2- (aq) + H3O+ (aq)
K1 = Ka (HClO2) = 1.1 ´ 10-2
NO2- (aq) + H3O+ (aq) « HNO2 (aq) + H2O (l)
For any reaction that can be written as the sum of two other reactions, K = K1K2. So, K = 24.2. HClO2 is the stronger acid (compared to HNO2) and NO2- is the stronger base (compared to ClO2-).
26.
a) From the chart on table 328, it is apparent that the pKa of cresol red is about 7.9 and the pKa of thymolphthalein is about 10.0. Therefore cresol red is a stronger acid than thymolphthalein; the conjugate base of cresol red must then be weaker than the conjugate base of thymolphthalein.b) If cresol red is red, then pH > 8.8. If thymolphthalein is colorless, the pH < 9.4. Hence, the pH is between 8.8 and 9.4.
27. This is a typical weak
acid problem. The reaction is (HA = HC9H7O4,
A- = C9H7O4-),
HA + H2O « A- + H3O+
33. This is a weak acid problem in reverse. The reaction between any weak acid and water is (HA = papH+):
The initial concetration of the acid is 0.072 M. We set up a table of equilibrium concentrations as follows:
HA A- H3O+ initial 0.072 0 0 change -x x x equilibrium 0.072 - x x x
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x = 4.65 ´ 10-3
This is a bit high for the assumption that 0.072 - x» 0.72. We can either solve using the quadratic equation of iterate. Since you know how to use the quadratic, let's iterate. Use the value of x you just estimated as a better approximation of the final concentration of the acid in the Ka expression as follows. Using a value of x = 4.7´ 10-3, we can write:
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x = 4.49 ´ 10-3
This is an "improved estimate of the equilibrium concentration of A- and H3O+. To insure that we have converged on an answer to two significant figures, which we are limited to in this problem, we say:
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x = 4.50 ´ 10-3
Since x = [H3O+], pH = -log (4.50 ´ 10-3) = 2.35
HA + H2O « A- + H3O+ In this case, you are given the concentration of the acid and the pH. Since this is an equilibrium reaction, a table of concentrations is helpful in defining quantities:
HA A- H3O+ initial 0.205 0 0 change -x x x equilibrium 0.205 - x x x In this case, x equals the hydronium ion concentration, which can be determined from pH as follows:
x = [H3O+] = 10-pH = 10-3.31 = 4.89 ´ 10-4
Substituting into the equilibrium expression yields:
35. This is a typical weak base problem. The reaction
is,
B + H2O « BH+ + OH-
The initial concetration of the base is 0.0667 M. We set up a table of equilibrium concentrations as follows:
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x = 2.31 ´ 10-4
In this case, the assumption that x is small relative to 0.0667 is good. Therefore, [OH-] = 2.31 ´ 10-4. Hydronium and hydroxide concentrations are related by Kw as follows.
[H3O+] [OH-] = Kw = 1.0 ´ 10-14
Therefore, [H3O+] = 4.33 ´ 10-11, and pH = 10.4.
43. Since "tris" is a base, its reaction with H3O+
(from HCl) will be:
tris + H3O+ ® trisH+ + H2O
The pKa of trisH+, the conjugate acid of tris is 8.08. (Why? Because, KaKb = Kw; taking the negative log of both sides gives pKa + pKb = pKw = 14.0.)To find the pH of buffer solutions, the Henderson-Hasselbach equation is helpful. To use it, you need to find the number of moles of the base and the conjugate acid. The base will react with the acid by the above reaction, and the number of moles before and after reaction are summarized below.
moles tris H3O+ trisH+ initial 0.050 0.025 0 change -0.025 -0.025 + 0.025 equilibrium 0.025 0 0.025
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45.
a) Very straight forward application of the Henderson-Hasselhoff equation:49.
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b) The reaction between acetic acid and hydroxide would be the formation of water and the conjugate base, acetate, or,
HA + OH- ® A- + H2O Thus, the amount of conjugate base will increase while the amount of acid will decrease as a result of the addition of hydroxide. Setting up a table to calculate the changes in the number of moles present gives,
moles HA OH- A- initial 0.050 0.010 0.020 change -0.010 -0.010 + 0.010 equilibrium 0.040 0 0.030
This is a Henderson-Humphrey problem in reverse. You have the pH and you know the pKa, so, you can treat the ratio of the base to acid as a single unknown. That is,
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Since you start with 0.0500 mol of formic acid, we can set up the following table, which summarizes the reaction of the acid with the added hydroxide ion.
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Solving for x in the above ratio yields,
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x = 0.0321
So, to get a pH of 4.00, 0.0321 mol of OH- needs to be added to the formic acid solution. Since the sodium hydroxide has a concentration of 0.0500 M, it will take 642 mL to deliver this amount.
55) To answer a question such as this fully, it helpful
to calculate the endpoint of the titration and to think about what are
the major species in solution at a given point in the titration. The equivalent
volume is the volume at which enough titrant has been added to completely
react with the analyte. In this case, the reaction is a simple neutralization
of base B with hydronium (from the strong acid titrant), yielding water
and the conjugate acid:
H3O+ (aq) + B (aq) ® H2O + BH+ (aq)
So, at the equivalent volume, the number of moles of H3O+ added will be equal to the number of moles of B present initially. Therefore:
This states that when 40.00 mL of HCl solution are added to the base solution, the base will be completely neutralized, but no excess acid has been added.
a) calculate the pH before any HCl is added.Since no acid has been added, this is a simple weak base problem. The reaction of the base with the water is:
B + H2O « BH+ + OH-
This is solved as any other equilibrium problem, that is, with a table of equilibrium concentrations.
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x = 8.01´ 10-3.
This is a bit high for the assumption that 0.1000 - x » 0.1000. To solve the problem by iteration, use the value of x you just estimated as a better approximation of the final concentration of the base in the Kb expression as follows. Using a value of x = 8.01 ´ 10-3, we can write:
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x = 7.68 ´ 10-3
A third iteration yields,
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x = 7.69 ´ 10-3 = [OH-]
To find the pH,
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pH = -log [H3O+] = 11.89
b) after 5.00 mL of HCl are added:As the neutralization progresses, base is being converted to conjugate acid by the reaction
H3O+ (aq) + B (aq) ® H2O + BH+ (aq)
Since the solution contains both the acid and the conjugate base, we can use the Henderson-Hasselbach equation. To keep track of the relative levels of the acid and conjugate base, it helpful to set up a table of moles; the number of moles of hydroxide is obtained by multiplying the volume by the concentration.
moles B H3O+ BH+ initial 4 x 10-3 0.5 x 10-3 0 change -0.5 x 10-3 -0.5 x 10-3 0.5 x 10-3 equilibrium 3.5 x 10-3 0 0.5 x 10-3 Using these values in the Henderson Hasselbach equation yields,
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pH = 11.66
Note that the pKa of the conjugate base is used and the form of the equation has the ration written as base/acid; this form works whether or not you are preparing a buffer from a base or an acid.
c) After 20.00 mL of HCl are added. This is also treated as a buffer problem:
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Using these values in the Henderson Hasselbach equation yields,
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pH = 10.81
d) After 39.90 mL of HCl are added: This is also treated as a buffer problem:
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Using these values in the Henderson Hasselbach equation yields,
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pH = 8.21
e) calculate the pH at 40.00 mL of HCl added (Veq).
At the equivalence point, all of the original base has been converted into its conjugate acid. To find the pH at this point, the reaction that needs to be considered is that between the conjugate acid and water. This is,HA + H2O « A- + H3O+
The equilibrium concentrations of these species are found as in any typical equilibrium problem. The initial concentration of the conjugate acid, however, is not the same as the initial concentration of the weak base. The number of moles of the conjugate base is the same as the initial number of moles of the weak acid (there is a 1:1 stoichiometric relationship), but the volume is greater because at the equilvalence point, 40.00 mL of acid solution have been added to the original volume. Thus the total volume is 80.00 mL, yielding a concentration of 0.0500 M.
concentration HA H3O+ A- initial 0.0500 0 0 change -x x x equilibrium 0.0500 -x x x
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x = 8.83 ´ 10-7.= [H3O+]
pH = -log [H3O+] = 6.05
f) calculate the pH at 40.10 mL of HCl added.Beyond the endpoint, the pH will be determined by the amount of excess hydronium that is added. To find the concentration, subtract the number of moles of acid that was consumed by the neutralization from the total number of moles added. Using the same type of table used in part b, we can write,
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The hydronium concentration is simply the moles of hydronium divided by the total volume (80.10 mL). So,
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pH = -log [H3O+] = 3.90
g) calculate the pH at 50.00 mL of HCl added.59)Using the same type of table used in part f, we can write,
moles B H3O+ BH+ initial 4.00 x 10-3 5.00 x 10-3 0 change -4.00 x 10-3 -4.00 x 10-3 4.00 x 10-3 equilibrium 0 1.00 x 10-3 4.00 x 10-3 So,
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pH = -log [H3O+] = 1.95
At the equivalence point, the moles of base present in an analyte solution is exactly equal to the number of moles of acid required to titrate the sample to the endpoint.
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The molar mass of diethylamine (B) is 73.13, thus there is a total of 0.0870 g of (C2H5)2NH in the 100 mL solution.
At the equivalence point all of the base has been converted to the conjugate acid; the total volume is 115.90 mL. The concentration of the conjugate acid is:
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To find the pH, use the Ka of the conjugate acid (Ka = Kw/Kb = 3.23 ´ 10-11). Setting up a table of equilibrium concentrations,
HA A- H3O+ initial 0.01027 0 0 change -x x x equilibrium 0.01027 - x x x
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x = 5.88 ´ 10-7 = [H3O+]
pH = -log [H3O+] = 6.23